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The markings on an aluminum ruler and a brass ruler begin at the left end; when the rulers are at \(0.00^{\circ} \mathrm{C}\), they are perfectly aligned. How far apart will the \(20.0 \mathrm{~cm}\) marks be on the two rulers at \(100.0^{\circ} \mathrm{C}\) if the left-hand ends are kept precisely aligned?

Short Answer

Expert verified
The 20.0 cm marks will be 0.006 cm apart at 100.0°C.

Step by step solution

01

Understand the Concept of Thermal Expansion

When materials are heated, they generally expand. The change in length of a material due to temperature change is determined by the formula: \( \Delta L = \alpha \cdot L_0 \cdot \Delta T \), where \(\Delta L\) is the change in length, \(\alpha\) is the coefficient of linear expansion, \(L_0\) is the original length, and \(\Delta T\) is the change in temperature.
02

Identify Known Values

We know the initial length \(L_0 = 20.0\, \mathrm{cm}\) for both rulers. The temperature change \(\Delta T = 100.0 - 0.0 = 100.0\, ^{\circ}\mathrm{C}\). The coefficient of linear expansion \(\alpha\) for aluminum is approximately \(22 \times 10^{-6} \, ^{\circ}\mathrm{C}^{-1}\) and for brass, it is approximately \(19 \times 10^{-6} \, ^{\circ}\mathrm{C}^{-1}\).
03

Calculate Expansion of the Aluminum Ruler

Using the formula \( \Delta L = \alpha \cdot L_0 \cdot \Delta T \), calculate the expansion for aluminum: \[ \Delta L_{aluminum} = 22 \times 10^{-6} \cdot 20.0 \cdot 100.0 = 0.044 \, \text{cm} \].
04

Calculate Expansion of the Brass Ruler

Similarly, calculate the expansion for brass: \[ \Delta L_{brass} = 19 \times 10^{-6} \cdot 20.0 \cdot 100.0 = 0.038 \, \text{cm} \].
05

Determine the Difference in Expansion

The difference in expansion between the two rulers is the difference in their respective expansions: \[ \Delta L_{difference} = 0.044 - 0.038 = 0.006 \, \text{cm} \].
06

Final Alignment at 100°C

The two rulers' 20.0 cm marks, originally aligned, will now be \(0.006\, \text{cm}\) apart at \(100.0^{\circ} \mathrm{C}\) due to their differential expansion.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Expansion Coefficient
When a material experiences a change in temperature, it can expand or contract. This behavior is described by the linear expansion coefficient, denoted as \[ \alpha \]. This coefficient is a crucial factor in the equation for thermal expansion. It measures how much a specific material will expand per degree temperature change. The formula for calculating the change in length (\( \Delta L \)) of a material is given by:\[ \Delta L = \alpha \cdot L_0 \cdot \Delta T \], where \( L_0 \) is the original length, and \( \Delta T \) is the change in temperature.
Different materials have unique \( \alpha \) values.
  • For aluminum, \( \alpha \) is approximately \( 22 \times 10^{-6} \, ^{\circ}\mathrm{C}^{-1} \).
  • For brass, \( \alpha \) is about \( 19 \times 10^{-6} \, ^{\circ}\mathrm{C}^{-1} \).
These coefficients imply aluminum expands more than brass for the same temperature change.
Understanding these coefficients can help predict and account for material expansion in engineering and construction projects.
Temperature Change Effect
Temperature change significantly influences the behavior of physical materials. When a material is exposed to increased temperatures, its molecules move more vigorously, causing the material to expand. This expansion due to temperature change is called thermal expansion. The change in length (\( \Delta L \)) directly relates to the extent of the temperature change (\( \Delta T \)).
  • As noted in the formula \( \Delta L = \alpha \cdot L_0 \cdot \Delta T \), a larger \( \Delta T \) results in a more significant change in length.
  • This relationship illustrates why precision instruments must consider temperature effects to avoid measurement errors.
In applications where materials are subject to large temperature variations, accommodating expansion is vital to prevent structural issues or inaccuracies.
Material Properties in Physics
Material properties, such as density, conductivity, and thermal expansion, influence how a substance behaves under different conditions. These properties are fundamental in physics and engineering. Each material's reaction to temperature changes varies due to its inherent properties.
Let's consider aluminum and brass, common materials with differing characteristics:
  • Aluminum, having a higher linear expansion coefficient, responds more to temperature changes compared to brass.
  • In our exercise, this difference means that at \(100.0^{\circ} \text{C}\), the 20.0 cm markings on the aluminum and brass rulers drift apart by \(0.006 \text{ cm}\).
Understanding material properties lets us anticipate the behavior of materials in everyday life and design solutions that consider their unique characteristics. This knowledge is crucial when choosing materials for specific uses, ensuring they function well within their operating conditions.

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Most popular questions from this chapter

(a) At what temperature do the Fahrenheit and Celsius scales give the same reading? (b) Is there any temperature at which the Kelvin and Celsius scales coincide?

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